Difference between revisions of "2020 AMC 10B Problems/Problem 22"
Advancedjus (talk | contribs) (→Solution 2) |
Kevinmathz (talk | contribs) |
||
Line 25: | Line 25: | ||
==Solution 2== | ==Solution 2== | ||
Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>. | Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>. | ||
+ | |||
+ | ==MAA Official Solution== | ||
+ | |||
+ | \begin{align*} | ||
+ | 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ | ||
+ | &= (2^{101} + 1)^2 - 2^{102} + 201\\ | ||
+ | &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. | ||
+ | \end{align*} | ||
==See Also== | ==See Also== |
Revision as of 23:41, 7 February 2020
Problem
What is the remainder when is divided by
?
Solution
Let . We are now looking for the remainder of
.
We could proceed with polynomial division, but the denominator looks awfully similar to the Sophie Germain Identity, which states that
Let's use the identity, with and
, so we have
Rearranging, we can see that this is exactly what we need:
So
Since the first half divides cleanly as shown earlier, the remainder must be ~quacker88
Solution 2
Similar to Solution 1, let . It suffices to find remainder of
. Dividing polynomials results in a remainder of
.
MAA Official Solution
\begin{align*} 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\
&= (2^{101} + 1)^2 - 2^{102} + 201\\ &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.
\end{align*}
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.